| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2005 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Single-piece PDF with k |
| Difficulty | Standard +0.3 This is a straightforward probability density function question requiring standard techniques: integrating to find k, using the pdf to find probabilities, and calculating E(X). The integration of (x-18)² is routine, and all three parts follow textbook procedures with no novel problem-solving required. Slightly easier than average due to the simple polynomial form. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\int_0^{24} \frac{(x-18)^2}{k} = 1\) | M1 | For equating to 1 and attempting to integrate |
| \(\left[\frac{(x-18)^3}{3k}\right]_0^{24} = 1\) | A1 | For correct integration with correct limits seen |
| \(\frac{2016}{k} - 1 \Rightarrow k = 2016\) | A1 (3) | For given answer legit obtained |
| (ii) \(p(x < 2) = \int_0^2 \frac{(x-18)^2}{2016} dx\) | M1 | For integration attempt between 0 and 2 (condone missing k) |
| \(= \left[\frac{(x-18)^3}{3 \times 2016}\right]_0^2\) | ||
| \(= \frac{(-16)^3 - (-18)^3}{6048}\) | ||
| \(= 0.2870(31/108)\) | A1 | For correct answer |
| Number of days \(= 0.287 \times 365 = 104\) or \(105\) | B1ft (3) | For multiplying their prob by 365 |
| (iii) mean \(= \int_0^{24} \frac{x(x-18)^2}{k} dx\) | M1 | For attempting to integrate xf(x) (condone missing k) |
| \(= \frac{1}{k}\left[\frac{x^4}{4} - \frac{36x^3}{3} + \frac{324x^2}{2}\right]_0^{24}\) | A1 | For one correct integrated term with correct limits(condone missing k) or For integration by parts correct first stage answer with limits seen(condone missing k) |
| A1 | For fully correct integrated expression with limits(condone missing k) | |
| \(= 5.14\left(5\frac{1}{7}\right)\) | A1 (4) | For correct answer |
**(i)** $\int_0^{24} \frac{(x-18)^2}{k} = 1$ | M1 | For equating to 1 and attempting to integrate
$\left[\frac{(x-18)^3}{3k}\right]_0^{24} = 1$ | A1 | For correct integration with correct limits seen
$\frac{2016}{k} - 1 \Rightarrow k = 2016$ | A1 (3) | For given answer legit obtained
**(ii)** $p(x < 2) = \int_0^2 \frac{(x-18)^2}{2016} dx$ | M1 | For integration attempt between 0 and 2 (condone missing k)
$= \left[\frac{(x-18)^3}{3 \times 2016}\right]_0^2$ | |
$= \frac{(-16)^3 - (-18)^3}{6048}$ | |
$= 0.2870(31/108)$ | A1 | For correct answer
Number of days $= 0.287 \times 365 = 104$ or $105$ | B1ft (3) | For multiplying their prob by 365
**(iii)** mean $= \int_0^{24} \frac{x(x-18)^2}{k} dx$ | M1 | For attempting to integrate xf(x) (condone missing k)
$= \frac{1}{k}\left[\frac{x^4}{4} - \frac{36x^3}{3} + \frac{324x^2}{2}\right]_0^{24}$ | A1 | For one correct integrated term with correct limits(condone missing k) or For integration by parts correct first stage answer with limits seen(condone missing k)
| A1 | For fully correct integrated expression with limits(condone missing k)
$= 5.14\left(5\frac{1}{7}\right)$ | A1 (4) | For correct answer7 The random variable $X$ denotes the number of hours of cloud cover per day at a weather forecasting centre. The probability density function of $X$ is given by
$$f ( x ) = \begin{cases} \frac { ( x - 18 ) ^ { 2 } } { k } & 0 \leqslant x \leqslant 24 \\ 0 & \text { otherwise } \end{cases}$$
where $k$ is a constant.\\
(i) Show that $k = 2016$.\\
(ii) On how many days in a year of 365 days can the centre expect to have less than 2 hours of cloud cover?\\
(iii) Find the mean number of hours of cloud cover per day.
\hfill \mbox{\textit{CAIE S2 2005 Q7 [10]}}